Suppose $b_1, b_2 \in l^2(Z_N)$. How to prove that the composition $T_{b_2} \circ T_{b_1} $ of the convolution operators $T_{b_2}$ and $T_{b_1}$ is the convolution operator $T_b$ with $b=b_2 \cdot b_1$? Hint use previous question where I asked: How to prove that convolution is associative? and also
$b_2 \cdot b_1(m) = \sum_{n=0}^{N-1} b_2(m-n)b_1(n)$.